System of Particles and Rotational MotionNEET Physics · Class 11 · NCERT Chapter 6

You can expect 2 to 4 questions from this chapter in NEET 2027. The chapter has high PYQ frequency. Moment of inertia of standard shapes, the parallel axis theorem, conservation of angular momentum and rolling motion (especially comparing sphere/cylinder/ring on inclines) are the most heavily tested concepts.

Memorise these seven: thin rod about its centre M L squared by 12, thin rod about its end M L squared by 3, ring about its diameter M R squared by 2 (and about its central axis M R squared), disc about its central axis M R squared by 2, solid sphere about a diameter 2 M R squared by 5, hollow sphere about a diameter 2 M R squared by 3, and hollow cylinder about its axis M R squared. NEET asks one of these almost every year.

The parallel axis theorem says that the moment of inertia about any axis I equals I_cm plus M d squared, where I_cm is the moment of inertia about a parallel axis through the centre of mass and d is the perpendicular distance between the two axes. NEET uses it to compute the moment of inertia of a body about an axis on the edge or far from the centre.

For a planar (flat) lamina lying in the xy plane, the moment of inertia about the z axis equals the sum of the moments about the x and y axes: I_z equals I_x plus I_y. NEET uses it to derive the moment of inertia of a ring or disc about a diameter from its known moment about the central axis.

For rolling without slipping, the kinetic energy splits between translational (one half M v squared) and rotational (one half I omega squared). The smaller the moment of inertia I, the more energy goes into translation and the higher the linear speed at the bottom. Solid sphere has I equals two fifths M R squared (small), so it reaches the bottom fastest. Ring has I equals M R squared (large), so it reaches last.

When the net external torque on a system is zero, angular momentum L equals I omega is conserved. Figure skater pulling arms in: I goes down (mass closer to axis), so omega goes up to keep L constant. Same logic for spinning planets and neutron stars formed from collapsing star cores.

For pure rolling without slipping, the velocity of the contact point with the ground is zero, which gives v_centre equals R times omega. If the body slides faster than it rotates, it skids. If it rotates faster than it slides, it spins. Rolling friction is static and does no work; kinetic friction (during slipping) does negative work and dissipates energy.